# How to understand the term $\frac{1}{-2E(\vec{p})} e^{-ip(x-y)}$ of Klein-Gordon propagator in Peskin & Schroeder's book?

I am reading Peskin & Schroeder's book on Chapter 2. I have a question about how to get the term $$\frac{1}{-2E(\vec{p})} e^{-ip(x-y)}$$. The original equation for propagator is

$$\langle 0 | [\phi(x), \phi(y)] | 0 \rangle = \int \frac{d^{3}p}{(2\pi)^3} \frac{1}{2E(\vec{p})} [ e^{-ip(x-y)} - e^{ip(x-y)} ].\tag{2.54}$$ We separate this equation, namely, $$\langle 0 | [\phi(x), \phi(y)] | 0 \rangle = \int \frac{d^{3}p}{(2\pi)^3} \frac{1}{2E(\vec{p})} e^{-ip(x-y)} + \int \frac{d^{3}p}{(2\pi)^3} \frac{1}{-2E(\vec{p})} e^{ip(x-y)}$$ Peskin & Schroeder's Book says that the energy $$p^{0}=-E(\vec{p})$$ in the second term is less than 0. I tried to expand $$e^{ip(x-y)}$$ in the following:

$$e^{ip(x-y)} = e^{i(p^{\mu}(x-y)_{\mu})} = e^{i(Et-\vec{p}\cdot(\vec{x}-\vec{y}))} = e^{i[-(-Et)-\vec{p}\cdot(\vec{x}-\vec{y})]}=e^{-i[p^{0}t+\vec{p}\cdot(\vec{x}-\vec{y})]}$$ However, it seems that I can not write $$e^{-i[p^{0}t+\vec{p}\cdot(\vec{x}-\vec{y})]}$$ as $$e^{-ip(x-y)}$$, where $$p^{0}=-E(\vec{p})$$. Where is the problem?

• In the second term $\vec p$ is a dummy variable of integration, so you can change variables from $\vec p$ to $-\vec p$. It doesn't affect the measure and it doesn't affect $E(\vec p)=|\vec p|^2 + m^2$.
– hft
Jul 27, 2022 at 4:24
• @hft Hi, Thanks. I understand. Jul 27, 2022 at 7:54

And because you are integrating over all momenta $$\vec{p}$$, for every $$\vec{x}-\vec{y}$$ you pass, you will also pass through a $$\vec{y}-\vec{x}$$. Meaning that effectively, and only inside the integral: $$$$\vec{x}-\vec{y}=\vec{y}-\vec{x}$$$$ and because this only happens in space components the time component of $$p$$ which is $$E$$ will gain a minus sign, telling you that you are dealing with "negative energy" particles (antiparticles):
$$\langle 0 | [\phi(x), \phi(y)] | 0 \rangle = \int \frac{d^{3}p}{(2\pi)^3} \frac{1}{2E(\vec{p})} e^{-ip(x-y)} \Big|_{p_0=E} + \int \frac{d^{3}p}{(2\pi)^3} \frac{1}{2(-E(\vec{p}))} e^{-ip(x-y)} \Big|_{p_0=-E}$$
Notice you could also have absorbed the minus in the time component of $$x$$ instead, telling you that those antiparticles can be thought as normal particles kind of travelling back in time: $$\langle 0 | [\phi(x), \phi(y)] | 0 \rangle = \int \frac{d^{3}p}{(2\pi)^3} \frac{1}{2E(\vec{p})} e^{-ip(x-y)} \Big|_{p_0=E, \ t_+} - \int \frac{d^{3}p}{(2\pi)^3} \frac{1}{2E(\vec{p})} e^{-ip(x-y)} \Big|_{p_0=E, \ t_-}$$ (This last paragraph has to be taken with caution, since in reality is more complex than that)